68-02 二叉搜索树的最近公共祖先

题目

对于有根树 T 的两个结点 p、q，最近公共祖先表示为一个结点 x，满足 x 是 p、q 的祖先且 x 的深度尽可能大（一个节点也可以是它自己的祖先）

例如，给定如下二叉搜索树: root = [6,2,8,0,4,7,9,null,null,3,5]

LeetCode

C++ 题解

根据BST特点，节点共三种情况：
p,q都在左子树（二者的值都小于根的值）
p,q都在右子树 （二者的值都大于根的值）
p,q分别在左右子树（此时最近公共祖先为root）

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) 
    {   
        if( root == nullptr )
            return nullptr;

        TreeNode* res;
        if( p->val < root->val && q->val < root->val )
            res = lowestCommonAncestor( root->left, p, q );
        else if( p->val > root->val && q->val > root->val )
            res = lowestCommonAncestor( root->right, p, q );
        else
            res = root;    
        
        return res;
     }
};

python 题解

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def lowestCommonAncestor(self, root, p, q):
        """
        :type root: TreeNode
        :type p: TreeNode
        :type q: TreeNode
        :rtype: TreeNode
        """
        if root == None:
            return None
        
        if root.val > p.val and root.val > q.val:
            res = self.lowestCommonAncestor(root.left,p,q)
        elif root.val < p.val and root.val < q.val:
            res = self.lowestCommonAncestor(root.right,p,q)
        else:
            res =root
        
        return res